Why Linearize Trigonometric Functions?
Trigonometric functions can be evaluated exactly at familiar angles such as \(0\), \(\pi/6\), \(\pi/4\), and \(\pi/3\). If an angle is close to one of these, a tangent-line model gives a simple way to estimate the function value without evaluating the function directly.
The key is to work with the angle in radians. In this convention, the derivative formulas take their standard forms: the derivative of \(\sin x\) is \(\cos x\), the derivative of \(\cos x\) is \(-\sin x\), and the derivative of \(\tan x\) is \(\sec^2 x\) wherever tangent is defined.
The base point \(a\) is an angle at which both the function value and derivative are known. The displacement \(x-a\) measures how far the angle of interest is from that base point. A useful linearization typically has a nearby target angle and a base point with simple exact trigonometric values.
The Three Trigonometric Linearizations
Substitute each trigonometric function and its derivative into the general linearization formula. For sine and cosine, the formulas apply at every real base angle. For tangent, the base angle must satisfy \(\cos a\ne 0\), since tangent is undefined where cosine is zero.
These are local estimates: they are intended for \(x\) near \(a\), not as identities valid for every \(x\). Notice the sign in the cosine formula. Its slope at \(a\) is \(-\sin a\), so the linear term is \(-\sin a\,(x-a)\).
Why the Linearization Works
The formula matches the function’s value and slope at the base point. It also gives a precise sense in which the model captures the function’s first-order behavior nearby. Here is the justification for any function \(f\) differentiable at \(a\).
At \(x=a\), the linearization gives \(L_a(a)=f(a)+f'(a)(a-a)=f(a)\).
Since \(L_a(x)=f(a)+f'(a)(x-a)\), its derivative is \(L_a'(x)=f'(a)\), so in particular \(L_a'(a)=f'(a)\).
For \(h\ne0\), subtract the linearization at \(a+h\) from the function value and divide by \(h\).
By differentiability at \(a\), the first term on the right tends to \(f'(a)\) as \(h\to0\). Therefore the whole right side tends to zero. Equivalently, the difference between the function and its linearization, divided by the angle displacement, tends to zero. This proves that the linearization captures the function’s first-order change near \(a\).
Worked Examples
Worked Example: Estimate Sine Near \(\pi/6\)
Estimate \(\sin(\pi/6+0.04)\), with the displacement measured in radians. Use \(a=\pi/6\). The exact base value and slope are \(\sin(\pi/6)=1/2\) and \(\cos(\pi/6)=\sqrt{3}/2\). Thus the sine linearization is:
At the target angle, \(x-\pi/6=0.04\). Substitute this displacement:
Rounded to five decimal places, the estimate is \(\sin(\pi/6+0.04)\approx 0.53464\). Check the model at its base: \(L_{\pi/6}(\pi/6)=1/2\), and its slope is the constant \(\sqrt3/2\), which equals \(\cos(\pi/6)\). The value and slope therefore match at the base angle.
Worked Example: Estimate Cosine Near \(\pi/3\)
Estimate \(\cos(\pi/3+0.02)\), with the displacement in radians. Choose \(a=\pi/3\). Since \(\cos(\pi/3)=1/2\) and \(\sin(\pi/3)=\sqrt3/2\), the cosine linearization is:
The target displacement is \(0.02\), so:
Rounded to five decimal places, the estimate is \(\cos(\pi/3+0.02)\approx 0.48268\). The model check gives \(L_{\pi/3}(\pi/3)=1/2=\cos(\pi/3)\). Its slope is \(-\sqrt3/2=-\sin(\pi/3)\), the correct cosine slope at the base. The negative sign predicts that cosine decreases for a small positive change from this angle.
Worked Example: Estimate Tangent Near \(\pi/4\)
Estimate \(\tan(\pi/4+0.01)\), where the displacement is in radians. Tangent is defined at \(a=\pi/4\), and \(\tan(\pi/4)=1\). Also, \(\sec^2(\pi/4)=1/\cos^2(\pi/4)=1/(1/2)=2\). Therefore:
Substitute the target displacement:
The estimate is \(\tan(\pi/4+0.01)\approx 1.02\). At the base angle, the model gives \(L_{\pi/4}(\pi/4)=1=\tan(\pi/4)\), and its slope is \(2=\sec^2(\pi/4)\). Both checks agree with the tangent function’s value and derivative there.
Using Degree Measurements Correctly
The linearization formulas use the numerical angle in radians, not degrees. Consider estimating \(\sin(31^\circ)\) from the familiar angle \(30^\circ=\pi/6\). The target is one degree above the base angle, but the displacement to insert in the formula is \(\pi/180\), not \(1\).
Here the numerical estimate is rounded to five decimal places. The model check is \(L_{\pi/6}(\pi/6)=1/2\), with slope \(\sqrt3/2\), as required. This example illustrates a general procedure: express the base and target angles in radians, calculate their radian difference, and then substitute that difference.
A Common Mistake: Treating Degrees as Radians
A tempting but incorrect calculation for the previous example is to use \(1\) as the displacement because the target is one degree above \(30^\circ\):
This is wrong because \(1\) here represents one radian, not one degree. The correct change is \(\pi/180\) radians, giving approximately \(0.51511\). The incorrect result is also a warning sign: sine values always lie between \(-1\) and \(1\), so a predicted sine value above \(1\) cannot be correct.
Another important limitation concerns tangent. The formula for \(\tan x\) can only be centered at an angle where tangent is defined. At \(\pi/2\), for example, \(\cos(\pi/2)=0\), so \(\tan(\pi/2)\) has no finite value and there is no tangent linearization at that base point. A nearby angle must also be chosen from the domain of tangent.
Choosing and Checking a Model
A good base angle is familiar, has a manageable derivative, and is close to the angle being estimated. After writing the appropriate formula, keep the displacement \(x-a\) explicit until substituting the target. That makes the sign of the change visible and helps prevent confusion between the base angle and the target angle.
For a quick check, evaluate the proposed linear function at \(x=a\) and differentiate it. The value should equal \(f(a)\), and the slope should equal \(f'(a)\). These checks catch common errors such as using \(\sin a\) instead of \(\cos a\) as the sine slope, or omitting the negative sign in the cosine linearization.
A linearization is a local estimate rather than an exact trigonometric identity. The examples use small angle changes, and the method’s justification explains why it captures the first-order behavior near the base point. The formula itself does not guarantee a specified numerical accuracy for an arbitrary distance from that point.
Check Your Understanding
Use radians unless a question explicitly gives degrees. For each estimate, identify the base angle and the displacement.
- Write the linearization of \(\sin x\) at \(a=\pi/4\).
- Use the linearization of \(\cos x\) at \(a=0\) to estimate \(\cos(0.03)\).
- Find the linearization of \(\tan x\) at \(a=\pi/6\), and state its slope.
- What radian displacement should be used when estimating \(\sin(46^\circ)\) from the base angle \(45^\circ\)?
- Why can tangent not be linearized at \(a=\pi/2\)?
